Past exam of the topics in applied mathematics course of the University of Oxford / 2009 / B / 5 / c / Solution

The fixed-point branches are $x=0$ and $x=\pm \sqrt{3-6/a}$ beginning at $a=2$. Zero is a saddle before that threshold and repelling afterwards. The nonzero branches are saddles until $a=3$, attracting for $3<a<7/2$, and repelling beyond their resonant unit-circle crossing. Solid portions in the diagram mark attracting states; dashed portions mark saddles or repellers.

At $a=3$, the two asymmetric two-cycles split from $x=\pm 1$ with components $[\pm \sqrt{3+3/a}\pm \sqrt{3-9/a}]/2$, pairing the two values with the same first sign. These branches are always unstable. In addition, the antipodal two-cycle occupies the constant component levels $x=\pm \sqrt{3}$ for every $a>0$, attracting only below $a=1/2$. It must be included for the actual printed map, even though the quadratic proposed in the previous part misses it. At a given parameter the two component levels represent one cycle, rather than two independent fixed points.